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Re: [wide-int] int_traits <tree>


On 10/17/2013 09:48 AM, Richard Biener wrote:
On Thu, 17 Oct 2013, Richard Sandiford wrote:

Richard Biener <rguenther@suse.de> writes:
On Thu, 17 Oct 2013, Richard Sandiford wrote:

Richard Biener <rguenther@suse.de> writes:
The new tree representation can have a length greater than max_len
for an unsigned tree constant that occupies a whole number of HWIs.
The tree representation of an unsigned 0x8000 is 0x00 0x80 0x00.
When extended to max_wide_int the representation is the same.
But a 2-HWI addr_wide_int would be 0x80 0x00, without the leading zero.
The MIN trims the length from 3 to 2 in the last case.
Oh, so it was the tree rep that changed?  _Why_ was it changed?
We still cannot use it directly from wide-int and the extra
word is redundant because we have access to TYPE_UNSIGNED (TREE_TYPE ()).
It means that we can now use the tree's HWI array directly, without any
copying, for addr_wide_int and max_wide_int.  The only part of decompose ()
that does a copy is the small_prec case, which is trivially compiled out
for addr_wide_int and max_wide_int.
"     2) addr_wide_int.  This is a fixed size representation that is
      guaranteed to be large enough to compute any bit or byte sized
      address calculation on the target.  Currently the value is 64 + 4
      bits rounded up to the next number even multiple of
      HOST_BITS_PER_WIDE_INT (but this can be changed when the first
      port needs more than 64 bits for the size of a pointer).

      This flavor can be used for all address math on the target.  In
      this representation, the values are sign or zero extended based
      on their input types to the internal precision.  All math is done
      in this precision and then the values are truncated to fit in the
      result type.  Unlike most gimple or rtl intermediate code, it is
      not useful to perform the address arithmetic at the same
      precision in which the operands are represented because there has
      been no effort by the front ends to convert most addressing
      arithmetic to canonical types.

      In the addr_wide_int, all numbers are represented as signed
      numbers.  There are enough bits in the internal representation so
      that no infomation is lost by representing them this way."

so I guess from that that addr_wide_int.get_precision is always
that "64 + 4 rounded up".  Thus decompose gets that constant precision
input and the extra zeros make the necessary extension always a no-op.
Aha.

For max_wide_int the same rules apply, just its size is larger.

Ok.  So the reps are only canonical wide-int because we only
ever use them with precision > xprecision (maybe we should assert
that).
No, we allow precision == xprecision for addr_wide_int and max_wide_int too.
But all we do in that case is trim the length.

Requiring precision > xprecision was option (5) from my message.

Btw, we are not using them directly, but every time we actually
build a addr_wide_int / max_wide_int we copy them anyway:

/* Initialize the storage from integer X, in precision N.  */
template <int N>
template <typename T>
inline fixed_wide_int_storage <N>::fixed_wide_int_storage (const T &x)
{
   /* Check for type compatibility.  We don't want to initialize a
      fixed-width integer from something like a wide_int.  */
   WI_BINARY_RESULT (T, FIXED_WIDE_INT (N)) *assertion ATTRIBUTE_UNUSED;
   wide_int_ref xi (x, N);
   len = xi.len;
   for (unsigned int i = 0; i < len; ++i)
     val[i] = xi.val[i];
}
Are you saying that:

      max_wide_int x = (tree) y;

should just copy a pointer?  Then what about:
No, it should do a copy.  But only one - which it does now with
the append-extra-zeros-in-tree-rep, I was just thinking how to
avoid it when not doing that which means adjusting the rep in
the copy we need to do anyway.  If fixed_wide_int_storage is
really the only reason to enlarge the tree rep.
      max_wide_int z = x;

Should that just copy a pointer too, or is it OK for the second constructor
to do a copy?

In most cases we only use the fixed_wide_int to store the result of the
computation.  We don't use the above constructor for things like:

      max_wide_int x = wi::add ((tree) y, (int) z);

wi::add uses y and z without going through max_wide_int.
Yes, I am aware of that.

What's the reason again to not use my original proposed encoding
of the MSB being the sign bit?  RTL constants simply are all signed
then.  Just you have to also sign-extend in functions like lts_p
as not all constants are sign-extended.  But we can use both tree
(with the now appended zero) and RTL constants representation
unchanged.
The answer's the same as always.  RTL constants don't have a sign.
Any time we extend an RTL constant, we need to say whether the extension
should be sign extension or zero extension.  So the natural model for
RTL is that an SImode addition is done to SImode width, not some
arbitrary wider width.
RTL constants are sign-extended (whether you call them then "signed"
is up to you).  They have a precision.  This is how they are
valid reps for wide-ints, and that doesn't change.

I was saying that if we make not _all_ wide-ints sign-extended
then we can use the tree rep as-is.  We'd then have the wide-int
rep being either zero or sign extended but not arbitrary random
bits outside of the precision (same as the tree rep).

As nearly 100% of all compile-time constant math is done on
the tree level pessimizing that just because we can invent sth
that only fits RTX and not tree looks wrong.  At least if we
can invent sth that fits _both_, which I think we can (see above).
This would be a disaster. the wide int rep is the same as the double-int rep. both do not have an explicit bit that says that this came from a signed or unsigned value. Once you go down that road then you have to start saying what you mean for every binary operation if the signedness does not match. And you have to say for every comparison what it means to compare unsigned numbers in a signed way......

Here is one place where double-int really did get it correct and we copied that.



The branch uses wide_int for RTL quite naturally, and changing it
wouldn't help do anything with the tree decompose routine.
Sure.  Changing it would not change anything for RTL (but the
wide-int-to-RTL converter maybe) but help tree decompose.

Oh yes, we'd need to sign-extend the values in lts_p and friends.
But that'll be only symmetric to the ltu_p and friend routines.

Richard.


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